CODEWITHSUNDEEP
c++functionreferencepass-by-reference
codekiddy
codekiddy

Reputation: 6137

passing reference to function

I have a code that does something like following example, and I'm not sure if this is correct because the executable runs as expected.

// source.cpp
void compute_x(int& ref)
{
    ref = 0;
}

void f(int x) 
{
    int local = x;
    local = 1;

    if (local)
    {
          return copute_x(local);
    }
    else return;
}

int main()
{
    f(2);
    return 0;
}

the code runs but, Is variable local valid once f returns?

Upvotes: 0

Views: 43

Answers (2)

anubhav deshwal
anubhav deshwal

Reputation: 136

No.the variable local is not valid as its scope is inbetween the braces of the f function defination ...after that It has no existence

Upvotes: 0

Anon Mail
Anon Mail

Reputation: 4770

The variable local goes out of scope after f returns.

After your edit: but the return value is returned from the function and subsequently returned from main.

Upvotes: 1

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